Integration by Parts

How to Integrate Products of Different Types of Functions

Reversing the Product Rule of Differentiation

The product rule for differentiating \(f(x)g(x)\) (i.e., \(\frac{d}{dx}(fg) = f'g + fg'\)) can be translated into the following rule for computing the antiderivative of a product:

\[\int fg' ~dx = fg - \int f'g ~dx\]

This rule is known as integration by parts and its usefulness relies on \(f'g\) being easier to integrate than \(fg'\). When applying integration by parts to the integral of a product, \(fg'\), the first consideration is determining which factor corresponds to \(f\) and which factor corresponds to \(g'\). This will be addressed below.

Notation

Integration by parts is often written in the following form

\[\int u ~dv = uv - \int v ~du.\]

where \(u = f(x)\), \(dv = g'(x) ~dx\), and therefore \(du = f'(x)~dx\) and \(v = g(x)\).

When to Apply Integration by Parts

The method of integration by parts is most commonly applied to integrating products of different types of functions. For example, the product of a polynomial and an exponential function or the product of a polynomial and a logarithmic function. Integration by parts can also be used to integrate inverse functions, like \(\ln(x)\).

LAE: How to Pick \(u\) and \(dv\)

When deciding which factor corresponds to \(u\) (i.e., \(f(x)\)), use the acronym LAE to help remember the order in which different types of functions are preferred.

• Logarithmic functions (e.g., \(\ln(x)\))

• Algebraic (including power, polynomial, and rational) functions (e.g., \(x^2\), \(1/x\), \(\sqrt{x}\))

• Exponential functions (e.g., \(e^x\))

In other words, a logarithmic function is our first choice for \(u\), if it appears in the integrand. If not, then our second choice is for an algebraic function, and our last choice would be an exponential function.

Once \(u\) has been selected, set \(dv\) to be the remaining factors of the integrand, including the differential \(dx\). Furthermore, if \(u = f(x)\) and \(dv = g(x) ~dx\), then

\[du = f'(x) ~dx ~~~~\text{and} ~~~~ v = \int g(x)~dx ~~~ \hbox{(no $+ C$ required)}\]

Example 1

Integrate the product of a power function and an exponential function

Compute \(\displaystyle \int 2x e^x ~dx.\)

Step 1:   Based on the integrand and the preferred LAE order, pick \(u\) and \(dv\).

Since the integrand is \(2xe^x\), the preferred choice for \(u\) is \(u=2x\) and therefore \(dv = e^x ~dx\).

Step 2:   Compute \(du\) and \(v\).

\[\begin{align*} u &= 2x & dv &= e^x~dx \\ du &= \frac{d}{dx}(2x) ~dx = 2~dx & v &= \int e^x~dx = e^x \end{align*}\]

Step 3:   Compute the integral using integration by parts.

\[\begin{align*} \int 2xe^x ~dx &= uv -\int v ~du \\ &= (2x)e^x - \int (e^x)2 ~dx\\ &= 2xe^x - 2\int e^x ~dx && \hbox{simplify}\\ &= 2xe^x - 2e^x + C \\ &= 2e^x(x-1) + C \end{align*}\]

Check Our Work.

We can verify our answer by showing \(2x e^x\) is the derivative of \(2e^x(x-1)\).

\[\begin{align*} \frac{d}{dx} 2e^x(x-1) &= 2e^x(x-1) + 2e^x && \text{product rule} \\ &= 2xe^x - 2e^x + 2e^x && \text{expand} \\ &= 2xe^x && \text{simplify} \end{align*}\]

Example 2

Integrate the product of a power function and a logarithmic function

Compute \(\displaystyle \int x^2 \ln(x) ~dx.\)

Step 1:   Based on the integrand and the preferred LAE order, pick \(u\) and \(dv\).

Since the integrand is \(x^2 \ln(x)\), the preferred choice for \(u\) is \(u=\ln(x)\) and therefore \(dv = x^2 ~dx\).

Step 2:   Compute \(du\) and \(v\).

\[\begin{align*} u &= \ln(x) & dv &= x^2~dx \\ du &= \frac{d}{dx}(\ln(x)) ~dx = \frac{1}{x}~dx & v &= \int x^2~dx = \frac{1}{3}x^3 \end{align*}\]

Step 3:   Compute the integral using integration by parts.

\[\begin{align*} \int x^2\ln(x) ~dx &= uv -\int v ~du \\ &= \ln(x)\left(\frac{1}{3}x^3\right) - \int \left(\frac{1}{3}x^3\right)\frac{1}{x} ~dx\\ &= \frac{1}{3}x^3\ln(x) - \frac{1}{3}\int x^2 ~dx && \hbox{simplify}\\ &= \frac{1}{3}x^3\ln(x) - \frac{1}{3}\cdot \frac{1}{3}x^3 + C\\ &= \frac{1}{3}x^3\ln(x) - \frac{1}{9}x^3 + C \end{align*}\]

Check Our Work.

We can verify our answer by showing \(x^2 \ln(x)\) is the derivative of \(\dfrac{1}{3}x^3\ln(x) - \dfrac{1}{9}x^3\).

\[\begin{align*} \frac{d}{dx} \left(\frac{1}{3}x^3\ln(x) - \frac{1}{9}x^3\right) &= \frac{1}{3}3x^2\ln(x) + \frac{1}{3}x^3\frac{1}{x} - \frac{1}{9}3x^2 && \text{product and sum rules} \\ &= x^2\ln(x) + \frac{1}{3}x^2 - \frac{1}{3}x^2 && \text{simplify}\\ &= x^2\ln(x) && \text{simplify} \end{align*}\]

Example 3

Integrating the logarithm

Compute \(\displaystyle \int \ln(x) ~dx.\)

Step 1:   Based on the integrand and the preferred LAE order, pick \(u\) and \(dv\).

Since the integrand is \(\ln(x)\), the preferred choice for \(u\) is \(u=\ln(x)\) and therefore \(dv = dx\).

Step 2:   Compute \(du\) and \(v\).

\[\begin{align*} u &= \ln(x) & dv &= 1~dx \\ du &= \frac{d}{dx}(\ln(x)) ~dx = \frac{1}{x}~dx & v &= \int 1~dx = x \end{align*}\]

Step 3:   Compute the integral using integration by parts.

\[\begin{align*} \int \ln(x) ~dx &= uv -\int v ~du \\ &= (\ln(x))x - \int x \frac{1}{x} ~dx\\ &= x\ln(x) - \int 1 ~dx && \hbox{simplify}\\ &= x\ln(x) - x + C \end{align*}\]

Check Our Work.

We can verify our answer by showing \(\ln(x)\) is the derivative of \(x\ln(x) - x\).

\[\begin{align*} \frac{d}{dx} \left(x\ln(x) - x\right) &= \ln(x) + x\frac{1}{x} - 1 && \text{product and sum rules} \\ &= \ln(x) + 1 - 1 && \text{simplify}\\ &= \ln(x) && \text{simplify} \end{align*}\]

Example 4

Area of a region

Compute the area of the region under the graph of \(\displaystyle f(x) = 3xe^{-2x} \) from \(x=0\) to \(x=4\).

Step 1:   Write the area of the region as a definite integral.

\[\hbox{Area} = \int_0^4 3xe^{-2x}~dx\]

Step 2:   Compute \(\displaystyle \int 3xe^{-2x} ~dx \) using integration by parts.

Pick \(u\) and \(dv\) and compute \(du\) and \(v\). (Recall \(\int e^{ax} ~dx = \frac{1}{a}e^{ax}+C\).)

\[\begin{align*} u &= 3x & dv &= e^{-2x}~dx \\ du &= \frac{d}{dx}(3x) ~dx = 3~dx & v &= \int e^{-2x}~dx = -\frac{1}{2}e^{-2x} \end{align*}\]

\[\begin{align*} \int 3xe^{-2x} ~dx &= uv - \int v ~du \\ &= (3x)\left(-\frac{1}{2}e^{-2x}\right) - \int \left(-\frac{1}{2}e^{-2x}\right)3 ~dx\\ &= -\frac{3}{2}xe^{-2x} + \frac{3}{2}\int e^{-2x} ~dx && \hbox{simplify}\\ &= -\frac{3}{2}xe^{-2x} + \frac{3}{2}\left(-\frac{1}{2}\right)e^{-2x} + C && \text{using } \int e^{ax}dx = \frac{1}{a}e^{ax}+C \\ &= -\frac{3}{2}xe^{-2x} - \frac{3}{4}e^{-2x} + C \\ \end{align*}\]

Check Our Work.

We can verify our answer by showing \(3xe^{-2x}\) is the derivative of \(-\dfrac{3}{2}xe^{-2x} - \dfrac{3}{4}e^{-2x}\).

\[\begin{align*} \frac{d}{dx} \left(-\frac{3}{2}xe^{-2x} - \frac{3}{4}e^{-2x}\right) &= -\frac{3}{2}e^{-2x} - \frac{3}{2}xe^{-2x}(-2) - \frac{3}{4}e^{-2x}(-2) && \text{product and sum rules} \\ &= -\frac{3}{2}e^{-2x} + 3xe^{-2x} + \frac{3}{2}e^{-2x} && \text{simplify}\\ &= 3xe^{-2x} && \text{simplify} \end{align*}\]

Step 3:   Evaluate \(\displaystyle \int_0^4 3xe^{-2x} ~dx \) using the answer to Step 2.

\[\begin{align*} \hbox{Area} &= \int_0^4 3xe^{-2x}~dx && \hbox{using Step 1}\\ &= \left(-\frac{3}{2}xe^{-2x} - \frac{3}{4}e^{-2x}\right) \Biggr|_0^4 && \hbox{using Step 2}\\ &= \left(-\frac{3}{2}(4)e^{-8} - \frac{3}{4}e^{-8}\right) - \left(-\frac{3}{2}(0)e^{0} - \frac{3}{4}e^{0}\right)\\ &= \left(-6e^{-8} - \frac{3}{4}e^{-8}\right) - \left(0 - \frac{3}{4}\right) && \hbox{since $e^0=1$}\\ &= \frac{3}{4} - \frac{27}{4}e^{-8} && \hbox{simplify} \end{align*}\]

Example 5

Evaluating a definite integral using integration by parts

Evaluate \(\displaystyle \int_1^e (4x+1)\ln(x) ~dx\).

Step 1:   Compute \(\displaystyle \int (4x+1)\ln(x) ~dx \) using integration by parts.

Pick \(u\) and \(dv\) and compute \(du\) and \(v\).

\[\begin{align*} u &= \ln(x) & dv &= 4x+1 ~dx \\ du &= \frac{d}{dx}\ln(x) ~dx = \frac{1}{x}~dx & v &= \int 4x+1~dx = 2x^2+x \end{align*}\]

\[\begin{align*} \int (4x+1)\ln(x) ~dx &= uv - \int v ~du \\ &= (\ln(x))(2x^2+x) - \int (2x^2+x)\frac{1}{x} ~dx\\ &= (2x^2+x)\ln(x) - \int 2x+1 ~dx && \hbox{simplify}\\ &= (2x^2+x)\ln(x) - (x^2+x) + C \\ &= (2x^2+x)\ln(x) - x^2 - x + C \end{align*}\]

Check Our Work.

We can verify our answer by showing \((4x+1)\ln(x)\) is the derivative of \((2x^2+x)\ln(x) - x^2 - x\).

\[\begin{align*} \frac{d}{dx} \left((2x^2+x)\ln(x) - x^2 - x\right) &= (4x+1)\ln(x) + (2x^2+x)\frac{1}{x} - 2x - 1&& \text{product and sum rules} \\ &= (4x+1)\ln(x) + 2x+1 - 2x - 1 && \text{expand}\\ &= (4x+1)\ln(x) && \text{simplify} \end{align*}\]

Step 2:   Evaluate \(\displaystyle \int_1^e (4x+1)\ln(x) ~dx\) using the answer to Step 1.

\[\begin{align*} \int_1^e (4x+1)\ln(x) ~dx &= \left((2x^2+x)\ln(x) - x^2 - x\right) \Big|_1^e && \hbox{using Step 1}\\ &= \left((2e^2+e)\ln(e) - e^2 - e\right) - \left(3\ln(1) - 2\right)\\ &= \left(2e^2+e - e^2 - e\right) - \left(0 -2\right) && \hbox{since $\ln(e)=1$, $\ln(1)=0$}\\ &= e^2 + 2 &&\hbox{simplify} \end{align*}\]

Example 6

An integral that requires two applications of integration by parts

Compute \(\displaystyle \int 5x^2e^x ~dx\).

Step 1:   Use integration by parts to compute \(\displaystyle \int 5x^2e^x ~dx\).

Pick \(u\) and \(dv\) and compute \(du\) and \(v\).

\[\begin{align*} u &= 5x^2 & dv &= e^x~dx \\ du &= \frac{d}{dx}(5x^2) ~dx = 10x~dx & v &= \int e^x~dx = e^x \end{align*}\]

\[\begin{align*} \int 5x^2e^x ~dx &= uv - \int v~du \\ &= 5x^2e^x - \int 10xe^x~dx \end{align*}\]

In order to compute the indefinite integral of \(10xe^x\), we need to use integration by parts again.

Step 2:   Use integration by parts to compute \(\displaystyle \int 10xe^x ~dx\).

Pick \(u\) and \(dv\) and compute \(du\) and \(v\).

\[\begin{align*} u &= 10x & dv &= e^x~dx \\ du &= \frac{d}{dx}(10x) ~dx = 10~dx & v &= \int e^x~dx = e^x \end{align*}\]

\[\begin{align*} \int 10xe^x ~dx &= uv - \int v ~du \\ &= 10xe^x - \int 10e^x~dx \\ &= 10xe^x - 10e^x + C \\ &= (10x - 10)e^x + C \end{align*}\]

Check Our Work.

We can verify our answer by showing \(10xe^x\) is the derivative of \((10x - 10)e^x\).

\[\begin{align*} \frac{d}{dx} (10x - 10)e^x &= 10e^x + (10x - 10)e^x && \text{product rule} \\ &= (10 + 10x - 10)e^x && \text{pull out common factor of $e^x$}\\ &= 10xe^x && \text{simplify} \end{align*}\]

Step 3:   Use the answer to Step 2 to complete the computations in Step 1.

\[\begin{align*} \int 5x^2e^x ~dx &= 5x^2e^x - \int 10xe^x~dx && \text{using Step 1}\\ &= 5x^2e^x - (10x - 10)e^x + C && \text{using Step 2}\\ &= (5x^2 - (10x - 10))e^x + C && \text{pull out common factor of $e^x$}\\ &= (5x^2 - 10x + 10)e^x + C && \text{simplify} \end{align*}\]

Check Our Work.

We can verify our answer by showing \(5x^2e^x\) is the derivative of \((5x^2 - 10x + 10)e^x\).

\[\begin{align*} \frac{d}{dx} (5x^2 - 10x + 10)e^x &= (10x - 10)e^x + (5x^2 - 10x +10)e^x && \text{product rule} \\ &= (10x - 10 + 5x^2 - 10x + 10)e^x && \text{pull out common factor of $e^x$}\\ &= 5x^2e^x && \text{simplify} \end{align*}\]